Moment of inertia of rod about an axis inclined

[haruspex]

Okay, got it.
The new length would be Lsinθ and we know the formula for axis perpendicular to rod as ML²/12. So now formula would be ML²sin²θ/12 .

According to your diagram and explanation at the start of the thread, it would be cos, not sin. But I wonder if you drew the diagram correctly. Shouldn't theta be the angle between the rod and the axis?

 

[Raghav Gupta]

According to your diagram and explanation at the start of the thread, it would be cos, not sin. But I wonder if you drew the diagram correctly. Shouldn't theta be the angle between the rod and the axis?

No, I corrected it in post 4 or 5 around. Yes theta is angle between rod and axis.
Thanks to you as well Haruspex.

 

[BvU]

I hate to interfere with all this lovely integration, but it really isn't necessary.
The contribution of a mass element to the MoI depends only on its distance from the axis. So you can squash the object parallel to the axis, as long as you preserve masses. In this case, you will have a shorter, fatter (or higher density) rod. The mass will not change. What will the new length be?

Haru is absoltely right. So if you have something to start with, this is a nice tool to take the most effective approach. In other cases, it's good to have something as basic as $$
I = \int r^2 dm$$ to start from.

And you can also see right through this ( I didn't until now) and realize that dm can be ... dx but also ... dy. And the ... dy work has already been done (if you know the formula or have a table at hand).